guias ejercicios matematicas 2 (departamento)
TRANSCRIPT
PROBLEMARIO RESUMEN
GUIAS PUBLICADAS POR EL DEPARTAMENTO DE
MATEMATICAS PURAS Y APLICADAS DE LA
UNIVERSIDAD SIMON BOLIVAR
TRIMESTRE: ENERO – MARZO 2008.
DISPONIBILIDAD
http://ma.usb.ve/cursos/
La guías a continuación corresponde a la semana 1,2,3,4,5,6,7,8
tiene las soluciones de las guías 1,2,4,5,6,7.
Sírvase de ayuda para practicar matemáticas 2.
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∫cosα(βx)dx = cosα−1(βx) sen(βx)
αβ+α−1
α
∫cosα−2(βx)dx0 L(")&7 C,--"
∫cos6(2x)dx0
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>
∫
√
x2−a2
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∫ √a − x2dx7 '()"#"/$%,4 x =
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∫
csc(x)dx =ln | csc(x) − cot(x)| + C0
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>
∫
dx√
x2+4x+5dx7 '()"#"/$%,4 (x + 2)2 + 1 = x2 + 4x + 5 : x + 2 = tan(t)0
!>
∫ √−x2 + x + 1dx7 '()"#"/$%,4 5
4− (x − 1
2)2 = −x2 + x + 1 : x − 1
2=
√
52
sen(t)0
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∫ √t2 − 6tdt7 '()"#"/$%,4 t − 3 = 3 sec(x)0
#>
∫
2x−1x2
−6x+13dx7 '()"#"/$%,4 2x−1
(x−3)2+4= (2x−6)+5
(x−3)2+40
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>
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sen(4y) cos(5y)dy0
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sen4(3t) cos4(3t)dt0
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π/4sen3(z)
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7 '()"#"/$%,4 1 + cos2(t) = 2 cos2(x) + sen2(x)7
2 cos2(x) + sen2(x) = cos2(t)(2 + tan2(t)) : 11+cos2(t)
= sec2(t)2+tan2(t)
0
(>
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x sen3(x) cos(x)dx7 '()"#"/$%,4 (3%-%$" %/3")#,$%8/ +&# +,#3"'0
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u = tan(x/2)9 −π < x < π 4#,/'7&#., $(,-I(%"# 7(/$%&/ #,$%&/,- *" '"/& 0 $&'"/& "/ (/,7(/$%&/ #,$%&/,- *" u1 J,#, +#&3,# +&# I("9 (',.&' -,' 7&#.(-,' *"- '"/& 0 *"- $&'"/& *"-,/)(-& *&3-"5
sen(x) = 2 sen(x/2) cos(x/2) = 2
(
u√
1 + u2
)(
1√
1 + u2
)
=2u
1 + u2
cos(x) = cos2(x/2) − sen2(x/2) =
(
1√
1 + u2
)2
−(
u√
1 + u2
)2
=1 − u2
1 + u2
K%/,-."/4"9 $&.& u = tan(x/2) "' x = 2 arctan(u)9 -(")& dx = 21+u2 1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
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1
(G
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0cos(x)dx
(1−sen2(x))(sen2(x)+1)21
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0 lımx→0 (cos(x))csc(x)"
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x1
√1+e−tdt
x"
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x−1"
% 0 lımx→∞
(
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x ln(x)"
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"
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@
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csc(5t)5
− t−5
2+1 + t2+1
2+1+ 5t2
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∫
(
5√
x + 23√
x2 + 12√
1−x2
)
dx2
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∫
(
5√
x + 23√
x2 + 12√
1−x2
)
dx =∫
5√
xdx + 2∫
3√
x2dx + 12
∫
dx√1−x2
= x15+1
1
5+1
+ 2x23+1
2
3+1
+ 12arc sen(x) + C.
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∫
√arctan(x)
1+x2 dx2
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7'%9
∫
√arctan(x)
1+x2 dx =∫ √
udu = u12+1
1
2+1
+ C
= arctan(x)12+1
1
2+1
+ C.
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!
∫ 3 sen2(x) cos(x)(7−sen3(x))2
dx"
*+,-./012 #$%&'()*+%($ ), -+./'$ () 0+*'+/,) t = 7 − sen3(x)1 2)%).$& 34) −dt =
3 sen2(x) cos(x)dx 5
∫
3 sen2(x) cos(x)
(7 − sen3(x))2dx = −
∫
dt
t2= − t−2+1
−2 + 1=
1
7 − sen3(x)+ C.
!!
∫
tan2(x)dx 6748)*)%-'+9 *)-4)*() 34) tan2(x) + 1 = sec2(x)!"*+,-./012
∫
tan2(x)dx =∫
(sec2(x) − 1)dx =∫
sec2(x)dx −∫
dx
= tan(x) − x + C.
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∫
y√
1 − ydy 6748)*)%-'+9 42','-) ), -+./'$ () 0+*'+/,) 1 − y = u!"
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∫
y√
1 − ydy = −∫
(1 − u)√
udu = −∫ √
udu +∫
u√
udu
= −u12+1
1
2+1
+∫
u1
2+1du
= −2u32
3+ 2u
32+1
5
= 2
(
− (1−y)32
3+ (1−y)
52
5
)
+ C
# !
∫
3x2
x2+1dx 6748)*)%-'+9 3x2 = 3x2 ± 3!"
*+,-./012
∫
3x2
x2+1dx =
∫
3x2+3−3x2+1
dx = 3∫
x2+1−1x2+1
dx
= 3∫
(
x2+1x2+1
− 1x2+1
)
dx = 3∫ (
1 − 1x2+1
)
dx
= 3(∫
dx −∫
dxx2+1
)
= 3x + 3 arctan(x) + C.
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2x − 2, x < 15, &B 1 < x < 2
3x2 − 2x x > 2*+,-./012
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∫
f(x)dx = F (x)+C !"#!$ F (x) =
x2 − 2x + C1, x < 15x + C2, %& 1 < x < 2
x3 − x2 + C3 x > 2'"# C,C1, C2, C3 ∈
R ()*$% +,$
−1 + C1 = 5 + C2
10 + C2 = 4 + C3
-) +,$ .)/) %$/ ,#) ./010(02) '"#(0#,) !$3$ %)(0%4)'$/5 lımx→a− F (x) = lımx→a+ F (x) =F (a)6 .)/) a = 1 7 a = 28
9#) ."%03*$ %"*,'0:# $%5 F (x) =
x2 − 2x + 6, x ≤ 15x, %& 1 < x < 2
x3 − x2 + 6 x ≥ 2
; f(x) =| 3x2 − 3 |8*+,-./012
∫
f(x)dx = F (x)+C !"#!$ F (x) =
x3 − 3x + C1, x ≤ −1−x3 + 3x + C2, %& −1 < x < 1x3 − 3x + C3 x ≥ 1
'"# C, C1, C2, C3 ∈
R ()*$% +,$
2 + C1 = −2 + C2
2 + C2 = −2 + C3
9#) ."%03*$ %"*,'0:# $%5 F (x) =
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<8 =)>) $* 3"%+,$?" !$ *) >/@A') !$ *) 4,#'0:# +,$ %$ !) $# $* 0#($/2)*" [a, b]B !$%.,C% !020!)[a, b] $# n %,30#($/2)*"% 0>,)*$%8 D)*',*$ $* @/$) !$* '"//$%."#!0$#($ ."*&>"#" '0/',#%'/0("
.)/) 2)/0"% 2)*"/$% !$ n En = 3, 4, 5, . . .;6 ."/ F*(01" G)>) n → ∞8
!; f(x) = x + 1B a = −1 7 b = 28*+,-./012
H"!$1"% )./"I01)/ $* @/$) 3)?" *) >/@A') y = x + 1 .)/) *"% −1 ≤ x ≤ 26 ,(0*0J)#!"."*&>"#"% '0/',#%'/0("% 7 '"#%0!$/)#!" '"1"5
∆xi = |xi − xi−1| =b − a
n=
3
n,∀i, 1 ≤ i ≤ n.
H)/) n = 36 A(R) ≈∑3
i=1 A(Ri) = 68
!"#$ % &'"%&'"()'*
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−1 −0.5 0 0.5 1 1.5 20
0.5
1
1.5
2
2.5
3Grafico de Y=X+1
!"#" n = 4$ A(R) ≈∑4
i=1 A(Ri) = 5, 6250%
−1 −0.5 0 0.5 1 1.5 20
0.5
1
1.5
2
2.5
3Grafico de Y=X+1
!"#" n = 5$ A(R) ≈ ∑5i=1 A(Ri) = 5, 4%
−1 −0.5 0 0.5 1 1.5 20
0.5
1
1.5
2
2.5
3Grafico de Y=X+1
!"#" n = 50$ A(R) ≈ ∑50i=1 A(Ri) = 4, 590%
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0.5
1
1.5
2
2.5
3Grafico de Y=X+1
!"#" n = k$ f(xi) = (−1 + 3ik) + 1 = 3i
k%
A(R) ≈k
∑
i=1
A(Ri) =k
∑
i=1
3
kf(xi) =
k∑
i=1
9i
k2=
9
k2
k∑
i=1
i =9
k2
k(k + 1)
2.
&'()*$ lımk→∞92
k(k+1)k2 = 4, 5+
, f(x) = 3x2 + x + 1- a = −1 % b = 1+*+,-./012
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∆xi = |xi − xi−1| =b − a
n=
2
n,∀i, 1 ≤ i ≤ n.
!"#" n = 3$ A(R) ≈ ∑3i=1 A(Ri) = 6, 4444+
−1 −0.5 0 0.5 1 1.5 20
5
10
15Grafico de Y=3X
2+X+1
!"#" n = 4$ A(R) ≈∑4
i=1 A(Ri) = 5, 75+
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0.5
1
1.5
2
2.5
3
3.5
4
4.5
5Grafico de Y=3X
2+X+1
!"#" n = 50$ A(R) ≈∑50
i=1 A(Ri) = 4, 1249%
−1 −0.5 0 0.5 10
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5Grafico de Y=3X
2+X+1
!"#" n = k$ f(xi) = 3(−1 + 2ik)2 + (−1 + 2i
k) + 1 = 12i2
k2 − 10ik
+ 3 &
A(R) ≈∑k
i=1 A(Ri) =∑k
i=12kf(xi) =
∑ki=1
24i2
k3 −∑k
i=120ik2 +
∑ki=1
6k
= 24k3
k(k+1)(2k+1)6
− 10k(k+1)2
+ 6
'()*+$ lımk→∞246
k(k+1)(2k+1)k3 − 20
2k(k+1)
k2 + 6 = 4% ,- .)/0#$ )1 2#)" )3"/4" .) 1" #)*056
-+78#)"." )- 4%
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0.5
1
1.5
2
2.5
3
3.5
4
4.5
5Grafico de Y=3X
2+X+1
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D
∫ sen(x)2−sen2(x)
dx2
!"#$%&'(
E5%-%F,/*& -, %)(,-*,* sen2(x) = 1 − cos2(x) &45"/".&' G(" 2 − sen2(x) = 1 + cos2(x)2H% #",-%F,.&' "- $,.4%& *" C,#%,4-" u = cos(x)9 du = − sen(x)dx2 >'%9
∫ sen(x)2−sen2(x)
dx =∫
−du1+u2 = − arctan(u) + C
= − arctan(cos(x)) + C.
!D
∫ sen(4x)cos(2x) cos(x)
dx2
!"#$%&'(
/$&/5#,# (/ $,.4%& *" C,#%,4-" ,+#&+%,*&9 /& '%".+#" "' "C%*"/5"2 7&# 5,/5&9 /&
*"4".&' &-C%*,# G(" "I%'5" -, +&'%4%-%*,* *" #"'&-C"# (/ +#&4-"., $&/ +#&+%"*,*"'
5#%)&/&."5#%$,'2 E5%-%F,/*& -, %)(,-*,* sen(2mx) = 2 sen(mx) cos(mx) &45"/".&' G("sen(4x) = 2 sen(2x) cos(2x) 0 sen(2x) = 2 sen(x) cos(x)2 >'%9
∫ sen(4x)cos(2x) cos(x)
dx =∫ 2 sen(2x) cos(2x)
cos(2x) cos(x)dx = 2
∫ sen(2x)cos(x)
dx
= 2∫ 2 sen(x) cos(x)
cos(x)dx = 4
∫
sen(x)dx
= −4 cos(x) + C.
"D
∫
cos3(3x) sen(3x)dx2 !"#$%&'(
∫
cos3(3x) sen(3x)dx =∫
cos2(3x) cos(3x) sen(3x)dx=
∫
(1 − sen2)(3x) cos(3x) sen(3x)dx
B",-%F,/*& "- $,.4%& *" C,#%,4-" u = 1 − sen2(3x) 0 du = −6 sen(3x) cos(3x)dx9 &45"/J".&' G("
∫
(1 − sen2)(3x) cos(3x) sen(3x)dx = −1
6
∫
udu =−u2
2+ C
=−(1 − sen2(3x))2
2+ C.
K2 L,--" -,' '%)(%"/5"' %/5")#,-"' *"8/%*,'6
!"#$ % &'"%&'"()'*
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!
∫ 0
−2g(t)dt "#$ g(t) = |t − 1| − 1%
*+,-./012
&#'# g(t) = |t − 1| − 1 =
t − 2 () t > 1
−t () t ≤ 1∫ 0
−2
g(t)dt =
∫ 0
−2
−tdt =−t2
2|0−2= 0 +
4
2= 2.
!!
∫ 4
11
w2 dw%*+,-./012
∫ 4
1
1
w2dw =
w−2+1
−2 + 1|41=
−1
w|41=
−1
4− −1
1=
3
4.
"!
∫ 2
−1(x − 2|x|)dx%
*+,-./012
&#'# f(x) = x − 2|x| =
3x () x < 0
−x () x ≥ 0∫ 2
−1
(x − 2|x|)dx =
∫ 0
−1
3xdx +
∫ 2
0
−xdx =3x2
2|0−1 +
−x2
2|20 =
−7
2.
#!
∫ π
2
0sen2(3x) cos(3x)dx%
*+,-./012
*+,-).,$/# +- ",'0)# /+ 1,2),0-+ r = sen(3x)3 dr = 3 cos(3x)dx 4 +- -)')5+ /+ )$5+62,")7$
(89+2)#2 + )$:+2)#2 (+2,$ 2+(9+"5)1,'+$5+; a = sen(0) = 0 4 b = sen(3π/2) = −1% <()3
∫ π
2
0
sen2(3x) cos(3x)dx =1
3
∫ −1
0
r2dr
=−1
3
∫ 0
−1
r2dr =−1
3
(
r3
3|0−1
)
=−1
9.
$!
∫ 3
−3
√
3− | t |dt%*+,-./012
=+,
f(x) =√
3− | t | =
√3 + t () t < 0
√3 − t () t ≥ 0
>,/# ?8+ f +( 8$, :8$")7$ 9,2 @9#2 (+2 f(−t) = f(t)!3 9#/+'#( 85)-).,2 +- A+#2+', /+
()'+52B,
∫ 3
−3
√
3− | t |dt = 2
∫ 3
0
√3 − tdt
5#',$/# +- ",'0)# /+ 1,2),0-+ u = 3 − t @du = −dt!3
∫ 3
0
√3 − tdt = −
∫ 0
3
√udu =
∫ 3
0
√udu
!"#$ % &'"%&'"()'*
!"###$
!" #$%&'(
∫ 3
−3
√
3− | t |dt = 2
∫ 3
0
√udu =
4u3/2
3|30 = 4
√3.
)
∫ π
−π(x5+ | sen(x) |)dx*
*+,-./012
+,-, f(x) =| sen(x) |=
− sen(x) "& −π ≤ x ≤ 0
sen(x) "& 0 ≤ x ≤ π+./'/-$01$( f $" 20/ 320%&40 5/' 6 x5
20/ 320%&,0 &-5/'* !01,0%$"(
∫ π
−π
(x5+ | sen(x) |)dx =
∫ π
−π
x5dx +
∫ π
−π
| sen(x) | dx
=x6
6|π−π + 2
∫ π
0
sen(x)dx
= 0 − 2 cos(x)|π0 = 2(1 + 1) = 4.
!)
∫ π/3
−π/3sen5(θ)dθ*
*+,-./012
7/ 320%&40 sen(θ) $" 20/ 320%&40 &-5/'( f(θ) = sen5(θ) 1/-8&$0 ., $"9 6/ :2$( f(−θ) =sen5(−θ) = − sen5(θ) = −f(θ)*
72$;,( 5,' $. <$,'$-/ #$ "&-$1'=/
∫ π/3
−π/3sen5(θ)dθ = 0*
*+,-./01 ',345163/76>
∫ π/3
−π/3
sen5(θ)dθ =
∫ π/3
−π/3
sen(θ)(sen2(θ))2dθ
=
∫ π/3
−π/3
sen(θ)(1 − cos2(θ))2dθ.
?1&.&@/0#, $. %/-8&, #$ A/'&/8.$ u = cos(θ)( du = − sen(θ)dθ %,0 a = cos(−π/3) = 1/26 b = cos(π/3) = 1/2* B"&(
∫ π/3
−π/3
sen5(θ)dθ =
∫ π/3
−π/3
sen(θ)(1 − cos2(θ))2dθ
= −∫ 1/2
1/2
(1 − u2)2du = 0.
")
∫ −1
−3t−2
(t2−4t+3)2dt*
*+,-./012
?1&.&@/0#, $. %/-8&, #$ A/'&/8.$ u = t2 − 4t+3( du = (2t− 4t)dt = 2(t− 2)dt %,0 a = 246 b = 8* B"&(
∫ −1
−3t−2
(t2−4t+3)2dt = 1
2
∫ 8
24duu2 = −1
2
∫ 24
8duu2
= −12
(
−1u
)
|248 = 12(− 1
24+ 1
8) = 1
24.
!"#$ % &'"%&'"()'*
!"###$
!" #$%%& %$ '&()*$'$ '& %$+ +),-)&./&+ 0-.1)2.&+3
4
∫ x3
√x
√t sen(t)dt"
*+,-./012
5&$
F (x) =∫ x3
√x
√t sen(t)dt =
∫ 0√
x
√t sen(t)dt +
∫ x3
0
√t sen(t)dt
= −∫
√x
0
√t sen(t)dt +
∫ x3
0
√t sen(t)dt
67%)1$.'2 &% 7()8&( 9&2(&8$ :-.'$8&./$% '&% ;<%1-%2= 2>/&.&82+ ?-&
Dx (F (x)) = −Dx
(
∫
√x
0
√t sen(t)dt
)
+ Dx
(
∫ x3
0
√t sen(t)dt
)
= −√
x2
sen(√
x) + 3x3√
x sen(x3).
!4
∫ x2
x(t + 1)dt"
*+,-./012
5&$
F (x) =∫ x2
x(t + 1)dt =
∫ 0
x(t + 1)dt +
∫ x2
0(t + 1)dt
= −∫ x
0(t + 1)dt +
∫ x2
0(t + 1)dt
67%)1$.'2 &% 7()8&( 9&2(&8$ :-.'$8&./$% '&% ;<%1-%2= 2>/&.&82+ ?-&
Dx (F (x)) = −Dx
(∫ x
0(t + 1)dt
)
+ Dx
(
∫ x2
0(t + 1)dt
)
= −(x + 1) + 2x(x2 + 1).
"4
∫ x2
xx2
tdt"
*+,-./012
5&$
F (x) =∫ x2
xx2
tdt =
∫ 0
xx2
tdt +
∫ x2
0x2
tdt
= −∫ x
0x2
tdt +
∫ x2
0x2
tdt
67%)1$.'2 &% 7()8&( 9&2(&8$ :-.'$8&./$% '&% ;<%1-%2= 2>/&.&82+ ?-&
Dx (F (x)) = −Dx
(
∫ x
0x2
tdt
)
+ Dx
(
∫ x2
0x2
tdt
)
= −x + 2x = x.
" #$%%& f′
(π2) +) f(x) =
∫ 3x
2xx2 sen(5t)dt"
*+,-./012
5&$
f(x) =∫ 3x
2xx2 sen(5t)dt =
∫ 0
2xx2 sen(5t)dt +
∫ 3x
0x2 sen(5t)dt
= −∫ 2x
0x2 sen(5t)dt +
∫ 3x
0x2 sen(5t)dt
!"#$ % &'"%&'"()'*
!"###$
!"#$%&'() *# "+$,*+ -*)+*,& ./'(&,*'0&# (*# 12#%/#)3 )40*'*,)5 6/*
f′
(x) = −Dx
(
∫ 2x
0x2 sen(5t)dt
)
+ Dx
(
∫ 3x
0x2 sen(5t)dt
)
= −2x2 sen(10x) + 3x2 sen(15x).
!5$3 f′
(π2) = −π2
2sen(5π)+ 3π2
4sen(15π
2)7 8&() 6/* sen(x+2kπ) = sen(x),∀k ∈ Z ⇒ sen(5π) =
sen(π) = −1 9 sen(15π/2) = sen(−π/2) = 07 -*'*,)5 6/*3 f′
(π2) = π2
27
7 :&##* #& $'0*;+&# (*<'$(& (* %&(& /'& (* #&5 5$;/$*'0*5 =/'%$)'*5 *' *# $'0*+>&#) 6/* 5*
$'($%&?
@ f(t) =
g(t), 5A −2 ≤ t < 0h(t), 5A 0 < t ≤ 1
0 5A ')
()'(* g(t) = −(t + 1)2 + 1 9 h(t) = |t − 1| + 17 B' *#
$'0*+>&#) [−2, 1]7*+,-./012
∫ 1
−2
f(t)dt = int0−2g(t)dt +
∫ 1
0
h(t)dt
(&() 6/* g(t) = −(t + 1)2 + 1 = −(t2 + 2t) 9 6/* h(t) = −t "&+& t ∈ [0, 1]3 0*'*,)5 6/*
∫ 1
−2f(t)dt = −
∫ 0
−2(t2 − 2t)dt −
∫ 1
0tdt = 2 − 1
2= 3
2
!@ f(x) =
1, 5A 0 ≤ x < 1x, 5A 1 ≤ x < 2
4 − x 5A 2 ≤ x ≤ 4*' *# $'0*+>&#) [0, 4]7
*+,-./012
∫ 4
0
f(x)dx =
∫ 1
0
dx +
∫ 2
1
xdx +
∫ 4
2
(4 − x)dx =9
2.
C7 :&##&+ *# 2+*& (* #& +*;$D' #$,$0&(& ")+ #&5 ;+2<%&5 (* #&5 5$;/$*'0*5 =/'%$)'*57
@ f(x) = x29 g(x) = 2 − x2
7
*+,-./012
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1
0
1
2
3
4
!"#$ % &'"%&'"()'*
!"###$
!"#" $"%$&%"# '% "#'" (' %" #')*+, %*-*."("/ ('0'-+1 2"%%"# 3#*-'#+ %+1 3&,.+1 3"#" %+1
$&"%'1 1' $&-3%' 4&' x2 = 1 − x25 61 ('$*#/ x = ±1/ ',.+,$'1 '% "#'" (' %" #')*+,
%*-*."(" '1
∫ 1
−1
(2 − x2)dx −∫ 1
−1
x2dx = 2
(∫ 1
0
(2 − 2x2)dx
)
= 2
(
2x − 2x3
3
)
|10 =8
3.
7 f(x) = x3/ g(x) = −x 8 x = 15
*+,-./012
−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4
−1.5
−1
−0.5
0
0.5
1
1.5
6% "#'" (' %" #')*+, %*-*."(" '1
∫ 1
0
x3dx +
∫ 1
0
xdx =
(
x4
4+
x2
2
)
|10 =3
4.
!7 f(x) = 1 − x2 − 2x/ g(x) = x2 + 2/ %+1 '9'1 $++#(',"(+1 8 %" #'$." x = 35*+,-./012
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
−20
−10
0
10
20
30
40
!"#" $"%$&%"# '% "#'" (' %" #')*+, %*-*."("/ ('0'-+1 2"%%"# 3#*-'#+ '% 3&,.+ :3+1*.*;+7
3"#" '% $&"% f(x) = 1 − x2 − 2x = 05 61.+ 1' 1".*1<"$' 3"#" x =√
2 − 1/ ',.+,$'1 '%
"#'" (' %" #')*+, %*-*."(" '1
A(R) =∫ 3
0(x2 + 2)dx −
∫
√2−1
0(1 − x2 − 2x)dx +
∫ 3√
2−1−(1 − x2 − 2x)dx
=∫ 3
0(x2 + 2)dx −
∫ 3
0(1 − x2 − 2x)dx =
∫ 3
0(2x2 + 2x + 1)dx
=(
2x3
3+ x2 + x
)
|30 = 18.
!"#$ % &'"%&'"()'*
!"###$
! f(x) =
{
(x + 1)2 + 2, "# x < 0(x − 1)2 + 2, "# x ≥ 0
$ g(x) = |x| + 1%
*+,-./012
−3 −2 −1 0 1 2 3−1
0
1
2
3
4
5
6
&'(' )'*)+*'( ,* '(,' -, *' (,./01 */2/3'-'4 -,5,20" 6'**'( 7(/2,(0 *0" 7+130" 7'(' *0"
)+'*," ", )+27*, 8+, −x + 1 = (x + 1)2 + 2 $ x + 1 = (x− 1)2 + 2% 9" -,)/(4 6'**'( *0" 7+130"7'(' *0" )+'*," x2 +3x+2 = 0 :x = −2 0 x = −1! $ x2−3x+2 = 0 :x = 2 0 x = 1!% 91301),",* '(,' -, *' (,./01 */2/3'-' ,"
A(R) =
(∫ −1
−2
(−x + 1)dx −∫ −1
−2
((x + 1)2 + 2)dx
)
+
(∫ 0
−1
((x + 1)2 + 2)dx −∫ 0
−1
(−x + 1)dx
)
+
(∫ 1
0
((x − 1)2 + 2)dx −∫ 1
0
(x + 1)dx
)
+
(∫ 2
1
(x + 1)dx −∫ 2
1
((x − 1)2 + 2)dx
)
A(R) =
∫ −1
−2
(−x2 − 3x − 2)dx +
∫ 0
−1
(x2 + 3x + 2)dx
+
∫ 1
0
(x2 − 3x + 2)dx +
∫ 2
1
(−x2 + 3x − 2)dx
;'-0 8+, *'" (,./01," "01 "/2,3(/)'"4
A(R) = 2
(∫ 1
0
(x2 − 3x + 2)dx +
∫ 2
1
(−x2 + 3x − 2)dx
)
= 2.
% <'*)+*,
∫ 4
2(4x + 3)dx )020 *#2/3, -, "+2'" -, =/,2'11 :>* 302'( *' 7'(3/)/?1 8+, -/@/-,
' [2, 4] ,1 n "+5/13,(@'*0" -, /.+'* *01./3+-4 ",*,))/01, ' xk )020 ,* ,A3(,20 /B8+/,(-0 -,
)'-' /13,(@'*0!%
*+,-./012
f(x) = 4x + 3 ," +1' C+1)/?1 )013/1+' ,1 [2, 4]4 ,1301)," f ," /13,.('5*, ,1 [2, 4]% D,' P = {a =
!"#$ % &'"%&'"()'*
!"###$
x0, x1, . . . , xn−1, xn = b} !"# $#%&'(')" %*+!,#% -*, '"&*%.#,/ [2, 4] -/"-* (#-# xk = 2+k∆x = 2+ 2kn
(/" k = 0, 1, . . . , n 0 ∆x = 2n1 2' 3*,*(('/"#4/3 xk = xk−15 f(xk) = f(xk−1) = 4xk−1 + 31 63 -*('%5
f(xk) = 4(2 + (k − 1)2
n) + 3 = 11 +
8(k − 1)
n.
73'5
∑nk=1 f(xk−1)∆x = 2
n
(
11n + 8n
(∑n
k=1 k) − 8)
= 2n
(
11n + 8n
n(n+1)2
− 8)
= 2n× (15n − 4)
= 30n−4n
.
8!*+/5
lımn→∞
n∑
k=1
f(xk−1)∆x = lımn→∞
30n − 4
n= 30
*, ,94'&* *:'3&*5 *"&/"(*3
∫ 4
2
(4x + 3)dx = lımn→∞
n∑
k=1
f(xk−1)∆x = 30.
;#%# #$/%&#% (!#,<!'*% 3!+*%*"('# / (/4*"&#%'/5 $/% =#./% *3(%'># # 4-'#3$#%?!3>1.*1
!"#$%&"'(' )"*+! ,-./#(%
0$1(%2(*$!2- '$ 3(2$*42"5(&
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B0 / -&' '%)(%"/3"' +#&2-".,' *%2(!" -, #")%&/ R ,$&3,*, +&# -,' )#5C$,' *" -,' "$(,$%&/"'
*,*,' D .("'3#" -, #"2,/,*, #"+#"'"/3,3%6,0 E"'+(;' "/$("/3#" "- 6&-(."/ *"- '7-%*& )"/=
"#,*& ,- 9,$"# )%#,# R "/ 3&#/& ,- "!" x0
@ y = x2
π, x = 4, y = 00
!"#$%&'(
0 0.5 1 1.5 2 2.5 3 3.5 40
1
2
3
4
5
6
F3%-%:,/*& "- .;3&*& *" *%'$&'4 V = π∫
4
0(x2
π)2dx = 1024
5π0
!@ y = 1
x, x = 2, x = 4, y = 00
!"#$%&'(
!"#$ % &'"%&'"()'*
!"###$
0 0.5 1 1.5 2 2.5 3 3.5 40
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
!"#$#%&'() *$ +,")() (* (#-.)-/ V = π∫
4
2( 1
x)2dx = π
40
1 y =√
9 − x2, y = 0 *'"2* x = −2 3 x = 30*+,-./012
−2 −1 0 1 2 30
0.5
1
1.5
2
2.5
3
!"#$#%&'() *$ +,")() (* (#-.)-/ V = π∫
3
−2(√
9 − x2)2dx = 116π30
0 4' $)- -#56#*'"*- 72)8$*+&- (#869* $& 2*5#)' R &.)"&(& 7)2 $&- 52:;.&- (* $&- *.6&.#)'*-
(&(&- 3 +6*-"2* $& 2*8&'&(& 2*72*-*'"&"#<&0 =*-76,- *'.6*'"2* *$ <)$6+*' (*$ ->$#() 5*'?
*2&() &$ @&.*2 5#2&2 R *' ")2') &$ *9* y0
!1 y =√
x, x = 0, y = 30*+,-./012
!"#$ % &'"%&'"()'*
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0 1 2 3 4 5 6 7 8 90
0.5
1
1.5
2
2.5
3
!"#$#%&'() *$ +,")() (* -&.-&/)'*.0 V = 2π∫
9
0(3 −
√x)xdx = 243π
51 2"/& 3).#4#$#(&(5
-)' *$ +,")() (* (#.-). V = π∫
3
0(y2)2dy1
6 x = 2√
y, x = 0, y = 41*+,-./012
0 0.5 1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
4
!"#$#%&'() *$ +,")() (* -&.-&/)'*.0 V = 2π∫
4
0(4 − (x2
4))xdx = 32π1 2"/& 3).#4#$#(&(5
-)' *$ +,")() (* (#.-). V = π∫
4
0(2√
y)2dy1
!6 x = y3/2, y = 9, x = 01*+,-./012
!"#$ % &'"%&'"()'*
!"###$
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
9
!"#$#%&'() *$ +,")() (* -&.-&/)'*.0 V = 2π∫
27
0x(9− x2/3)dx = 6561π
41 2"/& 3).#4#$#(&(5
-)' *$ +,")() (* (#.-). V = π∫
9
0(y3/2)2dy1
6 y = 4x, y = 4x21
*+,-./012
0 0.2 0.4 0.6 0.8 10
0.5
1
1.5
2
2.5
3
3.5
4
!"#$#%&'() *$ +,")() (* -&.-&/)'*.0 V = 2π∫
1
0x(4x − 4x2)dx = 2π
31 2"/& 3).#4#$#(&(5
-)' *$ +,")() (* &/&'(*$&. V = π∫
4
0((
√y
2)2 − (y
4)2)dy1
71 8'-9*'"/* *$ :)$9+*' (*$ .;$#() <*'*/&() &$ =&-*/ <#/&/ $& /*<#;' *' *$ 3/#+*/ -9&(/&'"*
&-)"&(& 3)/ $& -9/:& y2 = x35 $& /*-"& x = 4 > *$ *?* x5 *' ")/') &0
0 0.5 1 1.5 2 2.5 3 3.5 4−8
−6
−4
−2
0
2
4
6
8
!"#$ % &'"%&'"()'*
!"###$
! "# $%&'# x = 4(*+,-./012
)'*+*,#-./ %+ 01'/./ .% &#$/-%23 V = 2π∫
4
0(4 − x)
√x3dx = 1024π
35(
!! "# $%&'# y = 8(*+,-./012
0 0.5 1 1.5 2 2.5 3 3.5 4
−5
0
5
0
2
4
6
8
10
12
14
16
XY
Z
4- %+ 5$#6&/7 8/.%0/2 /92%$:#$ ;<% %+ 2/+*./ .% $%:/+<&*/- &<=/ :/+<0%- .%2%#0/2
&#+&<+#$ %2 +# 65<$# .% &/+/$ $/>/( )'*+*,#-./ %+ 01'/./ .% #$#-.%+#23 V = π∫
4
0
(
82 − (8 −√
x3)2
)
dx704π
5(
?( @%# R +# $%5*A- #&/'#.# 8/$ y = x2= y = x( 4-&<%-'$% %+ :/+<0%- .%+ 2A+*./ ;<% $%2<+'#
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
! 4+ %>% x(*+,-./012
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−1
0
1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
YX
Z
!"#$ % &'"%&'"()'*
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!"#$#%&'() *$ +,")() (* &-&'(*$&./ V = π∫
1
0(x2 − x4)dx = 2π
150
1 2$ *3* y0*+,-./012
!"#$#%&'() *$ +,")() (* 4&.4&-)'*./ V = 2π∫
1
0(x − x2)xdx = π
60
!1 5& -*4"& y = x0*+,-./012
6$ 7#-&- $& -*7#)' &4)"&(& 8)- y = x29 y = x &$-*(*()- (* $& -*4"& y = x: 8)(*+).
);.*-<&- =>* &$ #'"*-.*4"&- *$ .)$#() 4)' 4>&$=>#*- .*44#)' "-&'.<*-.&$ 8*-8*'(#4>$&-
'). (&-& >'& -*7#)' 4>9& &-*& ') *. '# 4>&(-&(&: '# "-#&'7>$&- ) -*()'(&0 ?*-): .@
4)'4#(*-&+). >'& .*44#A' "-&'.<*-.&$ (#&7)'&$ (* +&'*-& =>* $& -*7#)' #'"*-.*44#)'
-*.>$"&'"* .*& >'& 4#-4>'B*-*'4#&: "*'(-#&+). *$ &-*& (* >'& C7>-& 7*)+*"-#4& 4)')D
4#(&E .)$&+*'"* B&$"&-& *F8-*.&- .> -&(#) r *' B>'4#A' (* $& <&-#&;$* x0
G;.*-<&'() $& C7>-& &'"*-#)-: 8)(*+). (*(>4#- =>* r = H#8)"*'>.& ×cos(π/4) ()'(*
H#8)"*'>.& = x − x20 2'")'4*.: r =
√
2
2(x − x2)0 ?)- $) "&'"): A(x) = πr2 = π
2(x − x2)2
9
V =∫
1
0A(x)dx0 2. (*4#-: V =
∫
1
0
π2(x − x2)2dx = π
600
I0 5& ;&.* (* >' .A$#() *. $& -*7#A' &4)"&(& 8)- y = 1 − x29 y = 1 − x4
0 5&. .*44#)'*.
"-&'.<*-.&$*. (*$ .A$#(): =>* .)' 8*-8*'(#4>$&-*. &$ *3* x: .)' 4>&(-&().0 2'4>*'"-* *$ <)$D>+*' (*$ .A$#()0
*+,-./012
!"#$ % &'"%&'"()'*
!"###$
−1 −0.5 0 0.5 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1 y=1−x2
y=1−x4
V = 2∫
1
0A(x)dx! "#$ A(x) = l2(x) %#&'()*( +,( -. )(/01$ 2( -. &.'( 2(- '#-02# (' '03(4)0".
"#$ )('5("4# .- (6( y78 9- -.2# 2(- ",.2).2# ($ :,$"01$ 2( -. *.)0.&-( x ('4. 2.2# 5#)
l(x) = (1 − x4) − (1 − x2) = x2 − x48 ;'0! V = 2
∫
1
0(x2 − x4)2dx = 16
3158
TU
BO
S
y
x
∆ ∆∆∆x
Vn =
2 π πππ
R · H
· ∆ ∆∆∆
x
R
H
∆ ∆∆∆y
x
y
R
H
Vn =
2 π πππ
R · H
· ∆ ∆∆∆
y
DIS
CO
S
∆ ∆∆∆x
R
V =
π πππ R
2 · ∆ ∆∆∆
x
∆ ∆∆∆y
R
V =
π πππ R
2 · ∆ ∆∆∆
y
x
y
AR
AN
DE
LA
S
∆ ∆∆∆y
R
V =
π πππ (
R2 –
r2 )
· ∆ ∆∆∆
y
x
y
r
∆ ∆∆∆x
R
V =
π πππ
( R
2 –
r2 )
· ∆ ∆∆∆
x
r
x
y
Só
lid
o d
e r
evo
lució
n g
en
era
do
po
r u
n r
ecin
to p
lan
o a
l g
irar
alr
ed
ed
or
del
eje
OY
H
R
y
x
y =
f (
x )
Recin
to g
en
era
do
r
R
H
x
y
Só
lid
o d
e r
ev
olu
ció
n g
en
era
do
Pro
yecció
n s
ob
re e
l eje
OX
:
Po
r tu
bo
s:
( (((
) )))[ [[[
] ]]]� ���= === = ===
− −−−⋅ ⋅⋅⋅
π πππ= ===
Rx
0x
dx
xf
Hx
2V
x
x0
H
x0
H –
f (
x0 )
V0 =
2 π πππ
x0 [
H –
f (
x0 )
] ∆ ∆∆∆
x
∆ ∆∆∆x
x
y
x0
y0 =
f (
x0 )
H∆ ∆∆∆
x∆ ∆∆∆
x
y
Pro
yecció
n s
ob
re e
l eje
OY
:
Po
r d
isco
s:
( (((
) )))� ���= === = ===
− −−−
� ��� � ���� ���� ��� � ���� ���
π πππ= ===
Hy
0y
21
dy
yf
V
H
R
x
y =
f (
x )
y0
∆ ∆∆∆y
x0 =
f-1 (
y0 )
R
H
x
y
∆ ∆∆∆y
∆ ∆∆∆y
x0
V0 =
π πππ [
f–1(
y0 )
] 2
· ∆ ∆∆∆
y
Só
lid
o d
e r
evo
lució
n g
en
era
do
po
r u
n r
ecin
to p
lan
o a
l g
irar
alr
ed
ed
or
del
eje
OX
H
R
y
x
y =
f (
x )
H
R
y
x
Recinto generador
Sólido de revolución generado
Pro
yecció
n s
ob
re e
l eje
OX
:
V0 =
π πππ· [
H2 –
f2(x
0)
] ·
∆ ∆∆∆x
Po
r a
ran
de
las:
( (((
) )))[ [[[
] ]]]� ���= === = ===
� ��� � ���� ���� ��� � ���� ���
− −−−⋅ ⋅⋅⋅
π πππ= ===
Rx
0x
22
dx
xf
HV
x
y
x0
y0 =
f (
x0 )
H∆ ∆∆∆
x
R
H
x
x0
x0
y0 =
f (
x0 )
∆ ∆∆∆x
H
R
y
x
x0
∆ ∆∆∆x
Pro
yecció
n s
ob
re e
l eje
OY
:
Po
r tu
bo
s:
( (((
) )))� ���= === = ===
− −−−⋅ ⋅⋅⋅
π πππ= ===
Hy
0y
1d
yy
fy
2V
H
R
y
x
y =
f (
x )
∆ ∆∆∆y
y0
x0 =
f -1
( y
0 )
H
R
y
x
y0
x0
V0 =
2 π πππ
y0 ·
f-1 (
y0 )
· ∆ ∆∆∆
y
∆ ∆∆∆y
!"#$%&"'(' )"*+! ,-./#(%
0$1(%2(*$!2- '$ 3(2$*42"5(&
67%(& 8 91."5('(&
:!$%- ; 3(%<-= >??@
39;AAA> B6%(52"5(C &$*(!( D 8E- F B
!"#$%$%&' '()"#%*&' +,#, -, '".,/, 6 01& 72 3(4#" "- '%)(%"/5" .,5"#%,-6 37-$(-& *" 8&-(."/"'*" '9-%*&' *" #"8&-($%9/: ;(/$%&/"' "<+&/"/$%,-"': -&),#%5.%$,': =%+"#49-%$,' 0 '(' %/8"#','2
>2 ?,--" -,' *"#%8,*,' *" -,' '%)(%"/5"' ;(/$%&/"'6
@ z = x3 ln(x2) + (log7(πx + e))52
!"#$%&'(
dz
dx= 3x2 ln(x2) + 2x2 +
5π (log7(πx + e))4
ln(7)(πx + e).
!@ y =√
ex2 + e√
x2
2
!"#$%&'(
dy
dx=
x exp(x2)√
exp(x2)+ exp(x).
"@ y = 32x2−3x
2
!"#$%&'(
dy
dx= ln(3)(4x − 3)32x2
−3x.
#@ y =√
log10 (3x2−x)2
!"#$%&'(
dy
dx=
ln(3)(2x − 1)
2 ln(10)√
log10 (3x2−x)
.
$@ y = xπ+1 + (1 + π)x2
!"#$%&'(
dy
dx= (π + 1)
(
xpi + ln(π + 1)(π + 1)x−1)
.
% @ y = xx$&/ x > 02
!"#$%&'(
dy
dx= xx(1 + ln(x)).
!"#$ % &'"%&'"()'*
!"###$
! y = g(x)f(x)" #$%&'() *$+ g(x) , f(x) #&' -./+0+'1.)23+# , g(x) +# #.+4%0+ %&#.5.6)7
*+,-./012
8. g′
(x) = 0 ⇔ g(x) = C 1&' C > 0" +'5&'1+#
dy
dx=
g(x)f(x)(
f′
(x) ln(g(x)) + f(x)
g′ (x)
)
g′
(x) 6= 0
8.
(
f′
(x) ln(C))
Cf(x) g′
(x) = 0
!! ex+y = 4 + x + y 9-+0.6)-) .4%3.1.5)!7
*+,-./012
8$%&'()4&# *$+ y -+%+'-+ -+ x" -+0.6)'-& )42&# 3)-&# -+ 3) +1$)1.:' 1&' 0+#%+15&
) x7 ;# -+1.0"
d
dx
(
ex+y)
=d
dx(4 + x + y),
&25+'+4&# *$+
(
1 + y′)
ex+y = 1 + y′
ex+y + y′
ex+y = 1 + y′
ex+y − 1 = y′
(1 − ex+y)
⇒ dydx
= −1.
"! y = coth( arctanh(x))7*+,-./012
dy
dx= − csch2( arctanh(x))
1 − x2.
#! y = ln( arccosh(x))7*+,-./012
dy
dx=
1√1 − x2 arccosh(x)
.
$ ! y = 5 senh2(x) + x2 cosh(3x) − x arcsenh(x3)7*+,-./012
dy
dx= 10 senh(x) cosh(x) + 2x cosh(3x) + 3x2 senh(3x) − arcsenh(x3) − 3x3
√x2 + 1
.
7 <) 0+(.:' )1&5)-) %&0 y = e−x2
, y = 0, x = 0 , x = 2" #+ =)1+ (.0)0 1&' 0+#%+15& )3 +>+ y7
;'1$+'50+ +3 6&3$4+' -+3 #:3.-& -+ 0+6&3$1.:' 0+#$35)'5+ 96+0 +3 (0?@1&!7
!"#$ % &'"%&'"()'*
!"###$
−3 −2 −1 0 1 2 3
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Gráfica de y=exp(−x2)
*+,-./012
!" #$%&'( $)*+",-.,) /) $(,-$ "- $)#01. -'(,-/- -"$)/)/($ /)" )2) y *) 3+)*,$- ). "- *0#+0).,)
&#+$-
−2
−1
0
1
2
−2
−1
0
1
20
0.2
0.4
0.6
0.8
1
40 +,0"05-3(* )" 36,(/( /) '-*'-$(.)*7 V = 2π∫ 2
0x exp(−x2)dx8 9)-"05-./( )" '-3:0( /)
;-$0-:") u = x27 (:,).)3(* <+) V = π
∫ 4
0exp(−u)du = π (1 − e−4)8
8 !.'+).,$) )" %$)- /) "- $)#01. -'(,-/- =($ y = cosh(2x), y = 0, x = − ln(5) > x = ln(5)8*+,-./012
?- $)#01. -'(,-/- =($ y = cosh(2x), y = 0, x = − ln(5) > x = ln(5) y *) 3+)*,$- ). "-
*0#+0).,) &#+$-
−2 −1.5 −1 −0.5 0 0.5 1 1.5 20
2
4
6
8
10
12
14
!"#$ % &'"%&'"()'*
!"###$
! A = 2∫ ln(5)
0cosh(2x)dx = senh(2x)|ln(5)
0 = 2, 4"
" #$%%& %$' '()*(&+,&' (+,&)-$%&'
.
∫
x2x2
dx"
*+,-./012
∫
x2x2
dx =2x2
−1
ln(2)+ C.
!.
∫ 2 ln(x)x
dx"
*+,-./012
∫
2 ln(x)
xdx = ln2(x) + C.
".
∫ 1
0(103x + 10−3x) dx"
*+,-./012
∫ 1
0
(
103x + 10−3x)
dx =1
3 ln(10)
(
103 − 10−3)
.
#.
∫
(x + 3)ex2+6xdx"
*+,-./012
∫
(x + 3)ex2+6xdx =1
2ex2+6x + C.
$.
∫ 1
0e2x+3dx"
*+,-./012
∫ 1
0
e2x+3dx =1
2
(
e5 − e3)
.
% .
∫
6v+93v2+9v
dv"
*+,-./012
∫
6v + 9
3v2 + 9vdv = ln |3v2 + 9v| + C.
&.
∫
cot(θ)dθ"
*+,-./012
∫
cot(θ)dθ = ln | sen(θ)| + C.
'.
∫
sec(u) csc(u)du"
*+,-./012 /$01 2*& sec(u) csc(u) = 1sen(u) cos(u)
= sen2(u)+cos2(x)sen(u) cos(u)
= tan(u) + cot(u)
∫
sec(u) csc(u)du =
∫
tan(u)du +
∫
cot(u)du = − ln | cos(u)| + ln | sen(u)| + C.
!"#$ % &'"%&'"()'*
!"###$
!
∫
x coth(x2) ln(senh(x2))dx"
*+,-./012
#$%&'(%)*+ $& ,%-.'+ *$ /%0'%.&$ u = ln(senh(x2))1 2$ +.3'$)$ 45$∫
x coth(x2) ln(senh(x2))dx =1
2
∫
u2du =u3
6+ C.
6$/+&/$-+2 $& ,%-.'+ *$ /%0'%.&$ 0$%&'(%*+
∫
x coth(x2) ln(senh(x2))dx =ln3(senh(x2))
6+ C.
!!
∫ senh(2z1/4)4√
z3dz"
*+,-./012
#$%&'(%)*+ $& ,%-.'+ *$ /%0'%.&$ u = 2z1/41 2$ +.3'$)$ 45$
∫
senh(2z1/4)4√
z3dz = 2
∫
senh(u)du = 2 cosh(u) + C.
6$/+&/$-+2 $& ,%-.'+ *$ /%0'%.&$ 0$%&'(%*+
∫
senh(2z1/4)4√
z3dz = 2 cosh(2z1/4) + C.
" 7+)2'*$0$ &% 0$8'9) R 45$ 2$ -5$230% $) &% 2'85'$)3$ :850%" ;+0-5&$ 5)% ')3$80%& <%0% $&
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$& ->3+*+ 45$ 2$ ')*',%"
"! ?& $@$ x A%0%)*$&%2!"
*+,-./012
R = f(x) B r = g(x)1 $)3+),$2 V = π∫ b
a(f2(x) − g2(x)) dx"
#! ?& $@$ y A,%2,%0+)$2!"
*+,-./012
V = 2π∫ b
ax (f(x) − g(x)) dx"
!"#$ % &'"%&'"()'*
!"###$
! "# $%&'# x = a (&#)&#$*+%)!,
*+,-./012
V = 2π∫ b
a(x − a) (f(x) − g(x)) dx,
!! "# $%&'# x = b (&#)&#$*+%)!,
*+,-./012
V = 2π∫ b
a(b − x) (f(x) − g(x)) dx,
, -*+)./%$% 0# $%1.2+ R 34% )% 54%)'$# %+ 0# ).14.%+'% 614$#, 7*$540% 4+# .+'%1$#0 8#$# %0
9*045%+ /%0 )20./* 34% )% 1%+%$#&4#+/* )% :#&% 1.$#$ R #0$%/%/*$ /% 0# $%&'# /#/#; 4'.0.&%
%0 5<'*/* 34% )% .+/.&#,
"! =0 %>% y (#$#+/%0#)!,
*+,-./012
R = f(y) ? r = g(y); %+'*+&%) V = π∫ d
c(f 2(y) − g2(y)) dy,
#! =0 %>% x (&#)&#$*+%)!,
*+,-./012
V = 2π∫ d
cy (f(y) − g(y)) dy,
! "# $%&'# y = 3 (&#)&#$*+%)!,*+,-./012
V = 2π∫ d
c(3 − y) (f(y) − g(y)) dy,
!"#$%$%&' &($%&)*+"'
@, "# .+'%+)./#/ /%0 )*+./* )% 5./% %+ /%&.A%0%); %+ :*+*$ /% B0%>#+/$* C$#:#5 D%00 (EFG@H
EIJJ!; .+9%+'*$ /%0 '%0<K*+*, L. 0# 9#$.#&.2+ %+ 0# 8$%).2+ %) /% P 0.A$#) 8*$ 8401#/# &4#/$#H
/#; %+&4%+'$% 0# .+'%+)./#/ L %+ /%&.A%0%) %)
L = 20 log10(121, 3P )
!"#$ % &'"%&'"()'*
!"###$
!"#$%"&'% () *)'+)#+," %" () -'%.+," -/' $") 0)"1) 1% '/#2 ) 115 1%#+0%(%.3*+,-./012
P =10
L20
121, 3
!"&/"#%.4 () *)'+)#+," %" () -'%.+," -/' $") 0)"1) 1% '/#2 ) 115 1%#+0%(%. %. +5$)( )
4,6360 × 103(+0'). -/' -$(5)1) #$)1')1)3
63 7) 8)5"+&$1 M 1% $" &%''%8/&/ %" () !"#$# % &'"() * %.
M = 0, 67 log10(0, 37E) + 1, 46
1/"1% E %. () %"%'59) 1%( &%''%8/&/ %" 2+(/:)&&.;</')3 !"#$%"&'% () %"%'59) 1%( &%''%8/&/
1% 8)5"+&$1 3
*+,-./012
E =10
M−1,460,67
0, 37
%"&/"#%.4 () %"%'59) 1%( &%''%8/&/ 1% 8)5"+&$1 %. +5$)( ) 5,0171 × 1082+(/:)&&.;</')3
=3 7) >/'8$() 1% ?&+'(+"5 1+#% @$% -)') n 5')"1% -/1%8/. )-'/A+8)' n! = 1 × 2 × · · · × n -/'
n! ≈
√2πn
(n
e
)n
B)(#$(% 10! 1% 8)"%') %A)#&)4 ($%5/ 1% >/'8) )-'/A+8)1) 8%1+)"&% () >/'8$() )"&%'+/'3
*+,-./012
C% >/'8) %A)#&) 10! = 3 628 8003 D&+(+E)"1/ () >/'8$() 1% ?&+'(+"5 10! ≈ 3 598 7003
!"#$%&"'(' )"*+! ,-./#(%
0$1(%2(*$!2- '$ 3(2$*42"5(&
67%(& 8 91."5('(&
:!$%- ; 3(%<-= >??@
39;AAA> B6%(52"5(C &$*(!( @ B
!"#$%$%&' '()"#%*&' +,#, -, '".,/, 80 1(2#" "- '%)(%"/3" .,3"#%,-4 5/3")#,$%6/ +&# +,#3"'7,-)(/,' %/3")#,-"' 3#%)&/&.83#%$,' 9 '('3%3($%6/0
:0 ;".("'3#" -, %*"/3%*,*
sec(x) =sen(x)
cos(x)+
cos(x)
1 + sen(x)
9 *"'+(<' (3%-=-%$"-, +,#, *"*($%# -, >6#.(-,
∫
sec(x)dx = ln | sec(x) + tan(x)| + C.
!"#$%&'(
sen(x)cos(x)
+ cos(x)1+sen(x)
= cos(x) cos(x)+sen(x)(1+sen(x))(1+sen(x)) cos(x)
= sen(x)+(sen2(x)+cos2(x))(1+sen(x)) cos(x)
= (1+sen(x))(1+sen(x)) cos(x)
= 1cos(x)
= sec(x).
?'=7
∫
sec(x)dx =
∫
sen(x)dx
cos(x)+
∫
cos(x)dx
1 + sen(x)
#",-%@,/*& -&' $,.2%&' *" A,#%,2-"' u = cos(x)7 du = − sen(x)dx 9 v = 1 + sen(x)7 dv =cos(x)dx "/ -,' %/3")#,-"' ,/3"#%&#"'
∫
sec(x)dx =∫ sen(x)dx
cos(x)+
∫ cos(x)dx
1+sen(x)
=∫ −du
u+
∫
dvv
= − ln |u| + ln |v| + C
= − ln |cos(x)| + ln |1 + sen(x)| + C
= ln |sec(x) + tan(x)| + C
B0 A,-C"
∫ 2π
0x| sen(x)|1+cos2(x)
dx D'()"#"/$%,4 E,), -, '('3%3($%6/ u = x−π 9 *"'+(<' (3%-%$" +#&+%"*,*"'
*" '%."3#=,F0
!"#$ % &'"%&'"()'*
!"###$
*+,-./012 !"#$%&#'() "$ *#+,%) (" -#.%#,$" u = x − π/ du = dx/
∫ 2π
0x| sen(x)|1+cos2(x)
dx =∫ π
−π
(u+π)| sen(u+π)|1+cos2(u+π)
du
=∫ π
−π
(u+π)| sen(u)|1+cos2(u)
du
=∫ π
−π
u| sen(u)|1+cos2(u)
du +∫ π
−π
π| sen(u)|1+cos2(u)
du.
0# 12'*%3'
u| sen(u)|1+cos2(u)
"4 2'# 12'*%3' %+5#./ 5). $) 6#'6)/
∫ π
−π
u| sen(u)|1+cos2(u)
du = 07 8(%*%)'#$+"'6"/ $#
12'*%3'
| sen(u)|1+cos2(u)
"4 2'# 12'*%3' 5#.7 9'6)'*"4/
∫ 2π
0x| sen(x)|1+cos2(x)
dx =∫ π
−π
π| sen(u)|1+cos2(u)
du
= π∫ π
−π
| sen(u)|1+cos2(u)
du
= 2π∫ π
0sen(u)
1+cos2(u)du.
!"#$%&#'() "$ *#+,%) (" -#.%#,$" u = cos(x)/ du = − sen(x)dx: #4;/
2π∫ π
0x| sen(x)|1+cos2(x)
dx = 2π∫ π
0sen(u)
1+cos2(u)du
= −2π∫ −1
1du
1+u2
= 2π∫ 1
−1du
1+u2 = 2 (arctan(u))1−1
= 2π(
π4
+ π4
)
= π2.
<7 =#$$" $#4 4%>2%"'6"4 %'6">.#$"4
?
∫
ln(x)dx.*+,-./012@6%$%&#'() %'6">.#*%)' 5). 5#.6"4 A *)'4%(".#'() *)+) f(x) = ln(x) A g′(x) =dx/ 6"'"+)4 B2" f ′(x) = dx/x A g(x) = x: #4%/
∫
ln(x)dx = x ln(x) −∫
xdx
x= x ln(x) − x + C.
!?
∫
ln2(x)dx7*+,-./012 @6%$%&#'() %'6">.#*%)' 5). 5#.6"4 A *)'4%(".#'() *)+) f(x) = ln2(x) A
g′(x) = dx/ 6"'"+)4 B2" f ′(x) = 2 ln(x)/x A g(x) = x: #4%/
∫
ln2(x)dx = x ln2(x) − 2
∫
x ln(x)dx
x= x ln2(x) − 2(x ln(x) − x) + C.
"?
∫ tan(x)dx√sec2(x)−4
7
*+,-./012
!"#$ % &'"%&'"()'*
!"###$
∫
tan(x)dx√
sec2(x) − 4=
∫
sen(x)dx
cos(x)√
1−4 cos2(x)cos2(x)
=
∫
sen(x)dx√
1 − 4 cos2(x)
!"#$%&#'() "$ *#+,%) (" -#.%#,$" u = cos(x)/ du = − sen(x)dx0 1"'"+)2 34"∫
tan(x)dx√
sec2(x) − 4= −
∫
du√1 − 4u2
.
!"#$%&#'() "$ *#+,%) (" -#.%#,$" u = cos(θ)/2/ du = − sen(θ)/2dθ0 ),1"'"+)2 34"
−∫
du√1 − 4u2
=
∫ − sen(θ)dθ
2√
sen2(x)=
−1
2θ + C,
("-)$-%"'() $)2 *#+,%)2
∫
tan(x)dx√
sec2(x) − 4=
−1
2arc cos(2 cos(x)) + C.
5
∫
arctan(x)dx6*+,-./012 71%$%&#'() %'1"8.#*%)' 9). 9#.1"2: 2"# f(x) = arctan(x) ; g′(x) = dx0 #2%/f ′(x) = dx
1+x2 dx/ g(x) = x ;
∫
arctan(x)dx = x arctan(x) −∫
xdx
1 + x2.
!"#$%&#'() "$ *#+,%) (" -#.%#,$": u = 1 + x2; du = 2xdx6 <"'"+)2 34"/
∫
xdx
1 + x2=
∫
du
2u=
1
2ln(u) + C =
1
2ln(1 + x2) + C
=). 4$1%+)/
∫
arctan(x)dx = x arctan(x) − 1
2ln(1 + x2) + C
!5
∫
x3 arctan(x)dx.*+,-./012 >'1"8.#'() 9). 9#.1"2/ *)'2%(".#+)2 *)+) g′(x) = x3
; f(x) = arctan(x)dx"'1)'*"2 g(x) = x4/4 ; f ′(x) = 1/(x2 + 1)dx0 #2%/
∫
x3 arctan(x)dx = 14
(
x4 arctan(x) −∫
x4
x2+1dx
)
= 14
(
x4 arctan(x) −∫
(
x2 − x2
x2+1
)
dx)
= 14
(
x4 arctan(x) −∫
(
x2 − x2+1−1x2+1
)
dx)
= 14
(
x4 arctan(x) −∫
(
x2 − x2+1x2+1
+ 1x2+1
)
dx)
= 14
(
x4 arctan(x) − 13x3 − x + arctan(x)
)
+ C.
!"#$ % &'"%&'"()'*
!"###$
!
∫
ln(ln(x)) 1xdx"
*+,-./012 #$%&'( )* +$,'$&-*. u = ln(x)/ du = dx/x" 012/
∫
ln(ln(x))1
xdx =
∫
ln(u)du = u ln |u| − u + C = ln |x| ln | ln |x|| − ln |x| + C.
!!
∫
cos(ln(x))dx"*+,-./012 345*6,$7'84 9(, 9$,5*1. 1*$ f(x) = cos(ln(x)) : g′(x) = dx/ f ′(x) = − sen(ln(x))dx
x
: g(x) = x" 012/
∫
cos(ln(x))dx = x cos(ln(x)) +
∫
x sen(ln(x))dx
x.
345*6,$4)( 9(, 9$,5*1 (5,$ +*;. 1*$ f(x) = sen(ln(x)) : g′(x) = dx/ f ′(x) = cos(ln(x))dxx
: g(x) = x" <&5*4*%(1 =>*/
∫
cos(ln(x))dx = x cos(ln(x)) + x sen(ln(x)) −∫
x cos(ln(x))dx
x.
?1 )*7',/
2
∫
cos(ln(x))dx = x cos(ln(x)) + x sen(ln(x))∫
cos(ln(x))dx =1
2(x cos(ln(x)) + x sen(ln(x))) .
"!
∫
(x3 − 2x) exp(x)dx"*+,-./012 345*6,$4)( 9(, 9$,5*1. 1*$ f(x) = x3−2x : g′(x) = exdx/ f ′(x) = (3x2−2)dx: g(x) = ex
012/
∫
(x3 − 2x) exp(x)dx = (x3 − 2x)ex −∫
(3x2 − 2)exdx
= (x3 − 2x)ex + 2ex − 3
∫
x2exdx
@(, (5,( -$)(/ '45*6,$4)( 9(, 9$,5*1 -$ >-5'%$ '45*6,$-/ 1*$ f(x) = x2/ g′(x) = exdx/
f ′(x) = 2xdx : g(x) = ex" A*4*%(1 =>*/
∫
x2exdx = x2ex − 2
∫
xexdx.
345*6,$4)( 9(, 9$,5*1 (5,$ +*;/ 1*$ f(x) = x/ g′(x) = exdx/ f ′(x) = dx : g(x) = ex
∫
x2exdx = x2ex − 2
(
xex −∫
exdx
)
= x2ex − 2xex + 2ex.
012/
∫
(x3 − 2x) exp(x)dx = (x3 − 2x) exp(x) + 2ex − 3(
x2ex − 2xex + 2ex)
.
!"#$ % &'"%&'"()'*
!"###$
!
∫
e2x−e−2x
e2x+e−2xdx"
*+,-./012
∫
e2x − e−2x
e2x + e−2xdx =
∫
senh(2x)
cosh(2x)dx =
1
2ln(cosh(2x)) + C.
!!
∫
e3x√4−e6x
dx"
*+,-./012 #$%&'(%)*+ $& ,%-.'+ *$ /%0'%.&$ u = e3x1 du = 3e3xdx2 3$ 4'$)$ 56$
∫
e3x
√4 − e6x
dx =1
3
∫
du√4 − u2
=1
3arc sen(u/2) + C =
1
3arc sen(e3x/2) + C.
7" 8$%) A =∫
exp(sx) cos(tx)dx 9 B =∫
exp(sx) sen(tx)dx" :$-6$340$ 56$sB + tA = exp(sx) sen(tx) + C ;36<$0$),'%= >%&&$ sB 64'&'(%)*+ ')4$<0%,'?) @+0 @%04$3!"
*+,-./012
sB =
∫
exp(sx) sen(tx)dx
A)4$<0%)*+ @+0 @%04$3= 3$% f(x) = sen(tx) 9 g′(x) = sesxdx1 f ′(x) = t cos(tx)dx 9 g(x) = esx"
B3C1
sB =
∫
exp(sx) sen(tx)dx = esx sen(tx) − t
∫
exp(sx) cos(tx)dx = esx sen(tx) − tA.
D3 *$,'01 sB + tA = esx sen(tx)"
" :$-6$340$ 56$
∫
cosα(βx)dx = cosα−1(βx) sen(βx)αβ
+α−1α
∫
cosα−2(βx)dx" E6$<+1 >%&&$∫
cos6(2x)dx"
*+,-./012 A)4$<0%)*+ @+0 @%04$3= 3$% f(x) = cosα−1(βx)1 g′(x) = cos(βx)dx1 f ′(x) = −(α −1) cosα−2(βx) sen(βx)βdx 9 g(x) = 1
βsen(βx)" B3'1
∫
cosα(βx)dx =1
βcosα−1(βx) sen(βx) + (α − 1)
∫
cosα−2(βx) sen2(βx)dx.
:%*+ 56$ sen2(βx) = 1 − cos2(βx)1 3$ 4'$)$ 56$∫
cosα(βx)dx =1
βcosα−1(βx) sen(βx) + (α − 1)
∫
cosα−2(βx)(βx)dx − (α − 1)
∫
cosα(βx)dx.
D3 *$,'01
α
∫
cosα(βx)dx =1
βcosα−1(βx) sen(βx) + (α − 1)
∫
cosα−2(βx)(βx)dx
∫
cosα(βx)dx =1
αβcosα−1(βx) sen(βx) +
(α − 1)
α
∫
cosα−2(βx)(βx)dx.
E6$<+1
∫
cos6(2x)dx = 112
sen(2x) cos5(2x) + 56
∫
cos4(2x)dx" B@&',%)*+ *$ )6$/+ &% F+0-6&%
%)4$0'+01 4$)$-+3 56$
!"#$ % &'"%&'"()'*
!"###$
∫
cos6(2x)dx =1
12sen(2x) cos5(2x) +
5
6
(
1
8sen(2x) cos3(2x) +
3
4
∫
cos2(2x)dx
)
!"#$"%&'(
∫
cos6(2x)dx =1
12sen(2x) cos5(2x) +
5
6
(
1
8sen(2x) cos3(2x) +
3
4
(
1
2
(
x +sen(4x)
4
)))
+ C.
!"#$%&"'(' )"*+! ,-./#(%
0$1(%2(*$!2- '$ 3(2$*42"5(&
67%(& 8 91."5('(&
:!$%- ; 3(%<-= >??@
39;AAA> B6%(52"5(C &$*(!( D B
!"#$%$%&' '()"#%*&' +,#, -, '".,/, 90 1(2#" "- '%)(%"/3" .,3"#%,-4 ,-)(/,' %/3")#,-"' 3#%)&/&.53#%6$,'7 '('3%3($%8/ 3#%)&/&.53#%$, +,#, #,$%&/,-%9,# : $&.+-"3,$%8/ *" $(,*#,*&'0
;0 <,--" -,' '%)(%"/3"' %/3")#,-"' (3%-%9,/*& "- $,.2%& *" =,#%,2-" '()"#%*&4
>
∫
√
x2−a2
xdx7 '()"#"/$%,4 x = a sec(t)0
!"#$%&'(
∫
√
x2−a2
xdx = a
∫
tan2(t)dt = a (tan(t) − t) + C
=√
x2 − a2 − a arcsec(x/a) + C =√
x2 − a2 − a arc cos(a/x) + C.
!>
∫ arc sen(x)√(1−x2)3
dx7 '()"#"/$%,4 x = sen(t)0
!"#$%&'(
∫ arc sen(x)√(1−x2)3
dx =∫
t sec2(t)dt = t tan(t) + ln(cos(t)) + C
= arc sen(x) x√
1−x2ln(
√1 − x2) + C.
">
∫ √a − x2dx7 '()"#"/$%,4 x =
√a cos(t)0
!"#$%&'(
∫ √a − x2dx = −a
∫
sen(t)dt = −a2
(t − sen(2t)/2) + C
= −a2
(t − sen(t) cos(t)) + C
= −a2
(
arc cos(x/√
a) − x√
a−x2
a
)
+ C.
?0 /$("/3#"
∫
√
4−x2
xdx +&# ."*%& *" -, '('3%3($%8/ u =
√4 − x2
: +&# ."*%& *" (/, '('3%3($%8/
3#%)&/&."3#%$, @$&/="/%"/3">0 A"'+("' $&.+,#" '(' #"'(-3,*&'0 B"$("#*" C("
∫
csc(x)dx =ln | csc(x) − cot(x)| + C0 !"#$%&'( D", u =
√4 − x2 ⇒ 4 − u2 = x2
: udu = −xdx0 E'%7
∫
√
4−x2
xdx =
∫
x√
4−x2
x2 dx
= −∫
u2du4−u2 = u + ln(u−2
u+2) + C = u + ln
(
(u−2)2
4−u2
)
+ C
=√
4 − x2 + ln(
(√
4−x2−2)2
x2
)
+ C.
!"#$ % &'"%&'"()'*
!"###$
!"# "$#" %&'"( )* #+&%*,&-") +% .&-/*" '+ 0&#*&/%+ x = 2 sen(t)( dx = 2 cos(t)dt1 2)3(
∫
√
4−x2
xdx = 2
∫ cos2(t)sen(t)
dt
= 2∫
(csc(t) − sen(t)) dt
= 2 (ln | csc(x) − cot(x)| + cos(t)) + C
= ln(
(csc(x) − cot(x))2) + 2 cos(t) + C
= ln
(
(
2−√
4−x2
x
)2)
+√
4 − x2 + C.
4%&#&-+5$+( &-/&) )"%6.*"5+) )"5 *76&%+)1
81 9+)6+%0& %&) )*76*+5$+) *5$+7#&%+) 6$*%*,&5'" ."-:%+$&.*;5 '+ .6&'#&'")
<
∫
dx√
x2+4x+5dx( )67+#+5.*&= (x + 2)2 + 1 = x2 + 4x + 5 > x + 2 = tan(t)1
*+,-./012
∫
dx√x2 + 4x + 5
dx = ln
∣
∣
∣
∣
1√x2 + 4x + 5
+ (x + 2)
∣
∣
∣
∣
+ C.
!<
∫ √−x2 + x + 1dx( )67+#+5.*&= 5
4− (x − 1
2)2 = −x2 + x + 1 > x − 1
2=
√
52
sen(t)1*+,-./012
∫ √−x2 + x + 1dx =
5
8
(
arc sen
(
2x − 1√5
)
−(
2x − 1√5
)
2√
1 + x − x2
√5
)
+ C.
"<
∫ √t2 − 6tdt( )67+#+5.*&= t − 3 = 3 sec(x)1
*+,-./012
∫ √t2 − 6tdt =
(2t − 6)√
t2 − 6t
4− 9 ln
(
t − 3 +√
t2 − 6t)
2+ C.
#<
∫
2x−1x2
−6x+13dx( )67+#+5.*&= 2x−1
(x−3)2+4= (2x−6)+5
(x−3)2+41
*+,-./012
∫
2x − 1
x2 − 6x + 13dx = ln
(
x2 − 6x + 13)
+5
2arctan
(
x − 3
2
)
+ C.
?1 @&%%+ %&) )*76*+5$+) *5$+7#&%+)
<
∫
sen(4y) cos(5y)dy1*+,-./012
∫
sen(4y) cos(5y)dy =−1
18cos (9y) +
1
2cos(y) + C.
!"#$ % &'"%&'"()'*
!"###$
!
∫
sen4(3t) cos4(3t)dt"*+,-./012
∫
sen4(3t) cos4(3t)dt =−1
24sen3 (3t) cos5 (3t) − 1
48sen (3t) cos5 (3t)
+1
192sen (3t) cos3 (3t) +
1
128sen (3t) cos (3t) +
3x
128+ C.
!!
∫
tan4(x)dx"*+,-./012
∫
tan4(x)dx =1
3tan3(x) − tan(x) + x + C.
"!
∫
tan−3(x) sec4(x)dx"*+,-./012
∫
tan4(x)dx =−1
2csc2(x) + ln(tan(x)) + C.
#!
∫
csc3(y)dy"*+,-./012
∫
csc3(y)dy =−1
2csc(y) +
1
2ln |csc(y) − cot(y)| + C.
$ !
∫ π/2
π/4sen3(z)
√
cos(z)dz"
*+,-./012 #$%&'(%)*+ $& ,%-.'+ *$ /%0'%.&$ u = cos(z) du = − sen(z)dz
∫ π/2
π/4
sen3(z)√
cos(z)dz =
∫
√
2/2
0
(1 − u2)√
udu
=
(
2
3u3/2 − 2
7u7/2
)
√
2/2
0
= 0, 3115.
%!
∫ 3 sen(z)cos2(z)+cos(z)−2
dz"
*+,-./012
∫
3 sen(z)
cos2(z) + cos(z) − 2dz =
1
3(ln |cos(z) − 1| − ln |cos(z) + 2|) + C.
&!
∫
dt1+cos2(t)
1 234$0$),'%5 1 + cos2(t) = 2 cos2(x) + sen2(x)1
2 cos2(x) + sen2(x) = cos2(t)(2 + tan2(t)) 6 11+cos2(t)
= sec2(t)2+tan2(t)
"
*+,-./012
∫
dt
1 + cos2(t)=
√2
2arctan
(√2
2tan(t)
)
+ C.
'!
∫
x sen3(x) cos(x)dx1 234$0$),'%5 37'&',$ ')7$40%,'8) 9+0 9%07$2"*+,-./012
∫
x sen3(x) cos(x)dx =x
4sen4(x) +
1
16sen3(x) cos(x) +
3
32sen(x) cos(x) − 3x
32+ C.